Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $q = \dfrac{-4y - 28}{4y^2 + 36y + 56} \div \dfrac{2y^2 - 4y}{-4y^3 - 32y^2 - 48y} $
Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{-4y - 28}{4y^2 + 36y + 56} \times \dfrac{-4y^3 - 32y^2 - 48y}{2y^2 - 4y} $ First factor out any common factors. $q = \dfrac{-4(y + 7)}{4(y^2 + 9y + 14)} \times \dfrac{-4y(y^2 + 8y + 12)}{2y(y - 2)} $ Then factor the quadratic expressions. $q = \dfrac {-4(y + 7)} {4(y + 2)(y + 7)} \times \dfrac {-4y(y + 2)(y + 6)} {2y(y - 2)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac {-4(y + 7) \times -4y(y + 2)(y + 6) } { 4(y + 2)(y + 7) \times 2y(y - 2)} $ $q = \dfrac {16y(y + 2)(y + 6)(y + 7)} {8y(y + 2)(y + 7)(y - 2)} $ Notice that $(y + 2)$ and $(y + 7)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {16y\cancel{(y + 2)}(y + 6)(y + 7)} {8y\cancel{(y + 2)}(y + 7)(y - 2)} $ We are dividing by $y + 2$ , so $y + 2 \neq 0$ Therefore, $y \neq -2$ $q = \dfrac {16y\cancel{(y + 2)}(y + 6)\cancel{(y + 7)}} {8y\cancel{(y + 2)}\cancel{(y + 7)}(y - 2)} $ We are dividing by $y + 7$ , so $y + 7 \neq 0$ Therefore, $y \neq -7$ $q = \dfrac {16y(y + 6)} {8y(y - 2)} $ $ q = \dfrac{2(y + 6)}{y - 2}; y \neq -2; y \neq -7 $